A vertex cover of a graph is a set of vertices such that each edge of the graph is incident to at least one vertex of the set. Now given a graph with several vertex sets, you are supposed to tell if each of them is a vertex cover or not.

Input Specification:

Each input file contains one test case. For each case, the first line gives two positive integers N and M (both no more than 1), being the total numbers of vertices and the edges, respectively. Then M lines follow, each describes an edge by giving the indices (from 0 to N−1) of the two ends of the edge.

After the graph, a positive integer K (≤ 100) is given, which is the number of queries. Then K lines of queries follow, each in the format:

N​v​​ [ [

where N​v​​ is the number of vertices in the set, and ['s are the indices of the vertices.

Output Specification:

For each query, print in a line Yes if the set is a vertex cover, or No if not.

Sample Input:

10 118 76 84 58 48 11 21 49 89 11 02 454 0 3 8 46 6 1 7 5 4 93 1 8 42 2 87 9 8 7 6 5 4 2

Sample Output:

NoYesYesNoNo

代码：

#include 
     
      using namespace std;const int maxn = 1e5 + 10;int N, M, T, K;int x, vis[maxn];vector
      
        v[maxn];bool flag = false;int step = 0;int main() {    scanf("%d%d", &N, &M);    for(int i = 0; i < M; i ++) {        int st, en;        scanf("%d%d", &st, &en);        v[st].push_back(i);        v[en].push_back(i);    }    scanf("%d", &T);    while(T --) {        step = 0;        memset(vis, 0, sizeof(vis));        scanf("%d", &K);        for(int i = 0; i < K; i ++){            scanf("%d", &x);            for(int j = 0; j < v[x].size(); j ++) {                if(vis[v[x][j]] == 0) {                    vis[v[x][j]] = 1;                    step ++;                }            }        }        if(step == M) printf("Yes\n");        else printf("No\n");    }    return 0;}
      
     

　　标记边被覆盖的数量 判断数目是否符合 行吧 理解错题意可还行

FH